astronomy question Rigel?

M

Moses

New member
The bright star Rigel in the constellation of Orion has a temperature about 3 times that of the Sun, and a
luminosity 64,000 times that of the Sun. What then is Rigel's radius compared to the radius of the Sun?
 
The book value is 62 solar radii, but this isn't the right answer for you so I'll quickly go through the steps:

Stefan's law says that the luminosity of a star _per_unit_area_ is proportional to the temperature raised to the forth power.

3^4 = 81x the Sun's brightness per unit area.

Since Rigel is 64,000 times brighter than the Sun, the surface area must be 64,000/81 = 790 times that of the Sun. The surface area of a sphere is proportional to radius squared, so square rooting this area gives a radius of 28.1 times that of the Sun.

Why does our value differ so much from the book value? The surface temperature figure is well off. Rigel's surface temperature is approximately twice that of the Sun, not three times higher. Using a more accurate figure gives the "right" answer. Presumably this is deliberate to ensure you calculate the figure rather than look it up in a book.
 
28,1 times as big as the sun.

L= 4 pi r^2 * T^4

so for example take r=3 and T=5

That would give you a luminosity of 70685,8.
64000 * 70685,8 = 4,52389 *10^9 (new L)
Now solve the following equasion for r (radius) and take t=15 (3 times 5):

4,52389 * 10^9 = 4 * pi * r^2 * 15
r= 84,3274

now determine how many times the original r that is:

( 84,3274 ) / 3 = 28,1091 (factor)

There you go. Also check out the link, I used info from there.
Cheers,
Maik
 
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